STRUCTURE OF LEAD ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( September 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). In this photo I present the electromagnetic laws governing the nuclear structure, but a student of Einstein (Dr Th. Kalogeropoulos ) criticised my discovery of nuclear force and structure by believing that the nuclear structure is due to the invalid relativity. In fact, here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Lead (Pb) has four stable isotopes: Pb-204, Pb-206, Pb-207, Pb-208. Lead-204 is entirely a primordial nuclide and is not a radiogenic nuclide. The three isotopes lead-206, lead-207, and lead-208 represent the ends of three decay chains called the uranium series (or radium series), the actinium series, and the thorium series, respectively. These series represent the decay chain products of long-lived primordial U-238, U-235, and Th-232, respectively. However, each of them also occurs, to some extent, as primordial isotopes which were made in supernovae, rather than radiogenically as daughter products. The fixed ratio of lead-204 to the primordial amounts of the other lead isotopes may be used as the baseline to estimate the extra amounts of radiogenic lead present in rocks as a result of decay from uranium and thorium. The longest-lived radioisotopes are Pb-205 with a half-life of ~15.3 million years and Pb-202 with a half-life of ~53,000 years. Of naturally-occurring radioisotopes, the shortest half-life is Pb-210 with a half-life of 22.20 years. The structure of lead consists of 8 horizontal planes of opposite spins, including four additional deuterons with S= +2 and S =-2 which exist over and under the structure of 8 horizontal planes. However here the four deuterons do not form horizontal squares but an up horizontal line (+UHL) and a down horizontal line (-DHL). So all these nucleons of the 8 horizontal planes and the +UHL and the –DHL give S = 0 . Moreover several protons of such a structure provide ' 44 blank positions able to receive 44 extra neutrons with two bonds per neutron for constructing the heavier stable isotope, the Pb-208, with S =0. You can see in Fig-7d the 8 horizontal planes along with the +UHL with the two deuterons of Pb-208 in my published paper “Nuclear structure is governed by the fundamental laws of electromagnetism”. (See also the fourth figure of the bottom of the page).' The core of lead , the Pb-164, with 82 protons and 82 neutrons (even number) forms a structure of high symmetry giving the 4 stable isotopes. In general, the structure of Pb-164 (core) has S =0 with 8 horizontal planes of opposite spins and 2 horizontal lines. So the structure of the stable Pb-208 is based on the structure of Pb-164 in which 44 extra neutrons of opposite spins make two bonds per neutron able to overcome the large number of pp repulsions. On the other hand in the heavier unstable nuclides the more extra neutrons than those of the stable Pb-208 (in the absence of blank positions) make single bonds leading to the beta minus decay. ' ' ''' '''STRUCTURE OF Pb-178, Pb-180, Pb-182, Pb-184, Pb-186, Pb-188, Pb-190, Pb-192, Pb-194 Pb-196, Pb-198, Pb-200, Pb-202, Pb-204, Pb-206, Pb-207, Pb-208, Pb-210, Pb-212 AND Pb-214 This group of unstable nuclides including the stable structures of Pb-204, Pb-206, Pb-207 and Pb-208 is based on the structure of Pb-164 (core) with S =0. For example the unstable Pb-202 of even number of extra neutrons with S= 0 has 38 extra neutrons of opposite spins. These extra neutrons fill the blank positions and make two bonds per neutron, but their small number with respect to the large number of pp repulsions of long range cannot give sufficient binding energies to pn bonds for overcoming the pp and nn repulsions. However in the stable structures of Pb-204, Pb-206, Pb-207, and Pb-208, the greater number of extra neutrons gives sufficient binding energies to pn bonds for overcoming the repulsions. (See the stable Pb-208 of S =0 with 44 extra neutrons in Fig.7d of my published paper). Note that the stable Pb-207 of odd number of extra neutrons with S = -1/2 is based also on the same structure of the Pb-164 with S=0. In this case of 43 extra neutrons the one absent neutron belongs to the four extra neutrons of positive spins of the up horizontal line (+UHL). That is S = 0 - 1(+1/2) = -1/2 In Fig. 7d you can see the four extra neutrons formed by the protons of the up horizontal line. Whereas in the unstable structures of Pb-210, Pb-212, and Pb-214 with S=0 the more extra neutrons than those of the stable Pb-208 (in the absence of blank positions) make single bonds leading to the beta minus decay. STRUCTURE OF Pb-179, Pb-181, Pb-183, Pb-185, Pb-187, Pb-189, Pb-191, Pb-193, Pb-195, Pb-197, Pb-199, Pb-201, Pb-203 AND Pb-215 After a careful analysis I found that the structures of such unstable nuclides with odd number of extra neutrons are based on another structure of the Pb-164 (core) having S = -2 . In this case the one deuteron of the up horizontal line (Fig. 7d of my published paper) changes the spin from S = +1 to S = -1 giving S = -2, because it goes to the down horizontal line (-DHL) for making horizontal bonds with a deuteron of the down line. Under this condition the unstable Pb-179 with S = -5/2 of 15 extra neutrons has 7 extra neutrons of positive spins and 8 extra neutrons of negative spins. That is S = -2 + 7(+1/2) + 8(-1/2) = -5/2 Whereas the unstable Pb-205 with S = -5/2 of 41 extra neutrons has 20 extra neutrons of positive spins and 21 extra neutrons of negative spins. That is S = -2 + 20(+1/2) + 21(-1/2) = -5/2 This unstable nuclide of odd number of extra neutrons (41) is the longest-lived radioisotope with a half-life of ~15.3 million years, because it exists with 41 extra neutrons existing between the stable Pb-204 of even number of 40 extra neutrons and the stable Pb-206 of even number of 42 extra neutrons. STRUCTURE OF Pb-209, Pb-211, Pb-213, AND Pb-215 WITH S = +9/2 After a careful analysis I found that the structure of the above unstable nuclides is based on another structure of Pb-164 (core) having S = +4. In this case the two deuterons of -DHL change their spins from S = -2 to S = +2 giving S = +4. Particularly they go to +UHL, for making horizontal bonds with the two deuterons of the up horizontal line. (See in Fig.7d of my published paper the +UHL with the two deuterons of positive spins ). Under this condition the unstable Pb-209 with S = +9/2 of 45 extra neutrons has 23 extra neutrons of positive spins and 22 extra neutrons of negative spins. That is S = +4 + 23(+1/2) + 22(-1/2) = +9/2 Note that this rearrangement reduces the number of blank positions because the deuterons of the -DHL at the new positions fill the blank positions formed by the protons of the +UHL. Under this condition of absent blank positions some extra neutrons of Pb-209 make single bonds leading to the beta decay. Moreover the unstable Pb-215 with S = +5/2 of 51 extra neutrons has 24 extra neutrons of positive spins and 27 extra neutrons of negative spins. That is S = +4 +24(+1/2) + 27(-1/2) = +5/2 This unstable nuclide In the absent of blank positions has several extra neutrons which make single bonds leading to beta minus decay. Category:Fundamental physics concepts